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Thread: A Trig Application

  1. #1
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    A Trig Application

    I have this question to do for my next class:

    A gutter is to be made from a metal sheet of width 3 metres by bending up one third of the sheet on each side through an angle θ. How should the θ be chosen so that the gutter will carry the maximum amount of water?

    This is what I pictured it to be:

    However, after that, I can't figure out where to start.

    Could someone help me out? A full solution would be very much appreciated.
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  2. #2
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    I wish I could show you a drawing, but here goes:

    \ | | /
    \|___|/

    The angle sides and the bottom are all of length one. We want to maximize area which we can split into three parts:

    Triangular cross sectional areas on the sides of the gutter, and the middle square c-s area. For some angle, $\displaystyle \theta$, the height is $\displaystyle cos(\theta)$ and the top "length" is $\displaystyle sin(\theta)$. The area of such triangle is $\displaystyle 0.5sin(\theta)cos(\theta)$. The area of the square is $\displaystyle sin(\theta)$. Add up the areas to get:

    $\displaystyle sin(\theta)cos(\theta)+sin(\theta)$

    To maximize the area, we need to take the derivative w.r.t. theta and set it equal to zero.

    $\displaystyle cos^2(\theta)-sin^2(\theta)-sin(\theta) = 0$

    $\displaystyle \theta=\frac{\pi}{6}$
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  3. #3
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    Oh. So would that be the angle formed in the triangle cross sectional area?

    Am I looking for the angle of that entire angle though? So would my final answer be 2pi/3?
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  4. #4
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    Quote Originally Posted by ty2391 View Post
    Oh. So would that be the angle formed in the triangle cross sectional area?

    Am I looking for the angle of that entire angle though? So would my final answer be 2pi/3?
    Yes, that angle is for the triangle, so you are right that you would have to add $\displaystyle \frac{\pi}{2}$ to get an angle of: $\displaystyle \frac{2\pi}{3}$.
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